Timeline for Feynman diagram and uncertainty
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2016 at 20:30 | comment | added | AccidentalFourierTransform | @EmilioPisanty quite the contrary: insisting in pretending that there is a time direction in the diagram doesn't help. I guess we are gonna have to agree to disagree :-) I not only imply that the time direction of external legs has no meaning, but I reaffirm it. I hope one day people will stop using a time arrow in Feynman diagrams. There are different schools of thoughts about this, and maybe in the end your interpretation is more useful. But I am quite convinced that it is not... | |
| Dec 6, 2016 at 20:27 | comment | added | Emilio Pisanty | For sure it is not a frame-by-frame picture of any physical process, but pretending that the time direction of the external legs has no meaning (which is implied by your answer) doesn't help. | |
| Dec 6, 2016 at 20:25 | comment | added | AccidentalFourierTransform | @EmilioPisanty that depends on what one means by a Feynman diagram. It may be an S matrix element or a correlation function. For the latter, there are no in- and out-states and there are no axes. For the former, one may organise the in- and out-states by the convention that the particles to the left (or bottom) of the diagrams are incoming, and those to the right (or top) are out-going. But this doesnt mean that such an axis represents the flow of time, because the parts of the diagram dont represent what physically happens: it is not a frame-by-frame picture of the physical process. | |
| Dec 6, 2016 at 20:21 | comment | added | Emilio Pisanty | This isn't really true - there is a time axis (but not a space or momentum axis). If you re-organize the outgoing legs and you change in which time direction they flow out you might get the same amplitude, but you change the processes to which it is applicable. Also, note how (for a vertical time axis) no legs can exit the diagram horizontally? That's because the time axis does matter. | |
| Dec 6, 2016 at 18:05 | comment | added | AccidentalFourierTransform | @VictorVMotti I would call it a mnemonic rule. Nothing really moves to the past, but thinking that an antiparticle behaves as it were a particle moving backwards in time helps to remember some properties of particles and antiparticles, such as that they have opposite charge, opposite helicity, they are related by the time reversal operator, etc. | |
| Dec 6, 2016 at 18:02 | comment | added | user56963 | So anti particle moving to the past is also incorrect or a mere interpretation. | |
| Dec 6, 2016 at 17:55 | comment | added | AccidentalFourierTransform | @VictorVMotti rotate it. Or take one line and move its origin to any other point in the plane. Or take a vertex and move it up/down. Or take a line and bend it. Do whatever you want with the pieces. | |
| Dec 6, 2016 at 17:54 | vote | accept | CommunityBot | ||
| Dec 6, 2016 at 17:54 | comment | added | user56963 | Can you show what do you by mean move around the pieces of the diagram? Are you suggesting to rotate it? | |
| Dec 6, 2016 at 17:48 | comment | added | AccidentalFourierTransform | @VictorVMotti see my comment above: Wikipedia, as many other informal sources (e.g., many professors), insist in the lie to the children that a Feynman diagram represents a physical process. You will have people tell you that time flows from top to bottom or statements of the sort. But you will know better: that is false. A Feynman diagram is a device to organise Wick contractions, and nothing else. | |
| Dec 6, 2016 at 17:46 | comment | added | user56963 | So the vertical axe is not anything. But you see Wikipedia puts the horizontal axe as $t$ to say as if anti particles move backward? | |
| Dec 6, 2016 at 17:45 | comment | added | AccidentalFourierTransform | @ACuriousMind the problem is, many introductions to QFT insist in "time flowing from left to right" in Feynman diagrams, which is of no help, not even to a beginner. I have no idea why people perpetuate this. | |
| Dec 6, 2016 at 17:40 | comment | added | ACuriousMind♦ | This. If one wants a formal statement, a Feynman diagram is an isomorphism class of a certain kind of graphs, and these graphs are not to be thought to be embedded in anything. | |
| Dec 6, 2016 at 17:38 | history | answered | AccidentalFourierTransform | CC BY-SA 3.0 |